Diff for "BashFAQ/067"

Differences between revisions 4 and 7 (spanning 3 versions)

How can I trim leading/trailing white space from one of my variables?

There are a few ways to do this:

   #POSIX, but fails if the variable contains newlines
   read -r var << EOF
   $var 
   EOF

One can also achieve in bash using a herestring

   # Bash still fails if the variable contains a newline.
   read  -r x <<< "$x"

(note: using IFS=$' \t' read -d -r x partially fix the "problem" of the newlines but adds a trailing \n)

There's also a solution using [:glob:extglob]:

   # Bash
   shopt -s extglob
   x=${x##+([[:space:]])} x=${x%%+([[:space:]])}
   shopt -u extglob

This also works in KornShell, without needing the explicit extglob setting:
# ksh x=${x##+([[:space:]])} x=${x%%+([[:space:]])}
There are many, many other ways to do this, using sed for instance:
# POSIX, suppress the trailing and leading whitespace on every lines x=$(echo "$x" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')
These are not necessarily the best, but they're known to work.

-  ⇤ ← Revision 4 as of 2008-05-21 20:52:00 → 
  Size: 1195
  Editor: GreyCat
  Comment: clean up
+   ← Revision 7 as of 2008-07-16 07:29:57 → ⇥
  Size: 1107
  Editor: pgas
  Comment: some precision in case the variable contains newlines
-Deletions are marked like this.
+Additions are marked like this.
 Line 3:
-There are a few ways to do this -- none of them elegant.

First, the most portable way would be to use `sed`:
+There are a few ways to do this:
-Line 8:
+Line 6:
-   # POSIX
   x=$(echo "$x" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')
+   #POSIX, but fails if the variable contains newlines
   read -r var << EOF
   $var 
   EOF
 Line 12:
-One can also achieve the same goal using Bash builtins:
+One can also achieve in bash using a herestring
-Line 15:
+Line 14:
-   # Bash
   # Remove leading whitespace:
   while [[ $x = [[:space:]]* ]]; do x=${x#[[:space:]]}; done

   # And now trailing:
   while [[ $x = *[[:space:]] ]]; do x=${x%[[:space:]]}; done
+   # Bash still fails if the variable contains a newline.
   read  -r x <<< "$x"
-Line 23:
+Line 18:
-Of course, the preceding example is not optimal, because it removes one character at a time, in a loop (although it's good enough in practice for most purposes).  If you want something a bit fancier, there's a solution using [:glob:extglob]:
+(note: using IFS=$' \t' read -d '' -r x partially fix the "problem" of the newlines but adds a trailing \n)

There's also a solution using [:glob:extglob]:
-Line 39:
+Line 36:
-There are many, many other ways to do this.  These are not necessarily the best, but they're known to work.
+There are many, many other ways to do this, using sed for instance:  
{{{
   # POSIX, suppress the trailing and leading whitespace on every lines
   x=$(echo "$x" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')
}}}

These are not necessarily the best, but they're known to work.